Electrostatics Question 6
Question 6 - 25 January - Shift 2
Match List I with List II :
| A. | Gauss’s Law in Electrostat ics |
I. | $\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}$ |
|---|---|---|---|
| B. | Faraday’s Law |
II. | $\oint \overrightarrow{{}B} \cdot d \overrightarrow{{}A}=0$ |
| C. | Gauss’s Law in Magnetism |
III. | $\oint \vec{B} \cdot \vec{l}=\mu_0 i_C+\mu_0 \in_0 \frac{d \phi_E}{d t}$ |
| D. | Ampere- Maxwell Law |
IV. | $\oint \overrightarrow{{}E} \cdot d \overrightarrow{{}s}=\frac{q}{\in_0}$ |
Choose the correct answer from the options given below :
(1) A-IV, B-I, C-II, D-III
(2) A-I, B-II, C-III, D-IV
(3) A-III, B-IV, C-I, D-II
(4) A-II, B-III, C-IV, D-I
Show Answer
Answer: (1)
Solution:
Formula: Electric Flux
Gauss’s Law of electrostatic
$ \phi=\oint \overrightarrow{{}E} \cdot d \overrightarrow{{}s}=\frac{q}{\epsilon_0} $
Faraday’s law $\oint \vec{E} \cdot d \vec{l}=\frac{-d \phi_B}{d t}$
Gauss’s law of magnetism $\oint \vec{B} \cdot d \vec{A}=0$
Ampere’s Maxwell law
$ \oint \vec{B} \cdot d \vec{l}=\mu_0 i_C+\mu_0 \in_0 \frac{d \phi_E}{d t} $
Where $i_C$ : Conduction current
$\epsilon_0 \frac{d \phi_E}{dt}$ : Displacement current
